Question

Calculate the osmotic pressure of $0.5\%$ solution of glucose (molecular mass 180) at ${18}^{o}C$. The value of solution constant is 0.0821 $litre-atm{K}^{-1}mo{l}^{-1}$.

• A. 0.6628 atm
• B. 1.6628 atm
• C. 0.2842 atm
• D. 0.1428 atm

Answer 1

Answer: (A)

0.6628 atm

Let's assume the mass of solution = 100g, and density of water = 1 g/ml

so, mass of glucose = $\frac{0.5}{100}\times 100=0.5g$

we have,

$\pi =iCRT$

since, i=1 for gulucose

$\pi =CRT$

$\pi =\frac{0.5}{180\times 100}\times 1000\times 291\times 0.0821$

We get,

$\pi =0.6628atm$

Hence, option (A) is correct.